The trace of the filter on a big subset

Let $$\scr{F}$$ be free filter ($$\cap\scr{F}=\emptyset$$) on a countable set $$X$$ and $$B\in\scr{F}$$. We define the trace of $$\scr{F}$$ on $$B$$ as follows $$\mathscr{F}_B=\{Y\cap B:~Y\in\scr{F}\}$$. $$\scr{F}$$ and $$\mathscr{F}_B$$ are isomorphic in the case of Frechet filter or in the case of any ultrafilter. Are $$\scr{F}$$ and $$\mathscr{F}_B$$ isomorphic in all cases ?

1 Answer

The answer is negative. You can get any infinite $$B$$ with infinite complement. Then let $$\scr{N}$$ be the Frechet filter on $$B$$ and $$\scr{F}$$ be the filter on $$X$$ generated by $$\scr{N}$$. Then $$\mathscr{F}_B = \scr{N}$$ and $$\mathscr{F}\not\simeq \mathscr{F}_B$$ because for any bijection $$\phi:X\to B$$ the set $$\phi(B)$$ has infinite complement and thus $$\phi(B)\notin\mathscr{F}_B$$.

But if $$B$$ has infinite complement in $$X$$ and has subset $$B\supset B_0\in\scr{F}$$ such that $$B-B_0$$ is infinite, then $$\mathscr{F}\simeq \mathscr{F}_B$$.